166
Section 6.4 - Logarithmic Functions
Objective #1:
Definition of a Logarithmic Function
Let’s work with solving some exponential equations. Recall exponential
functions are one-to-one with a domain of (– ∞, ∞) and a range of (0, ∞).
We want to use these facts to solve the following.
Solve the following:
Ex. 1a 4x = 16
1
Ex. 1d 2x =
Ex. 1b (1/3)x = 243
Ex. 1e 6x = 0
8
Ex. 1c 5x = 625
Ex. 1f 5x = – 25
Solution:
a)
Since 16 = 42, then x = 2.
b)
Since (1/3)x = 3 – x = 243 and 243 = 35, then 3 – x = 35. Thus,
€
– x = 5 which means x = – 5.
c)
Since 625 = 54, then x = 4.
1
1
d)
Since = 3 = 2 – 3, then x = – 3.
8
e)
2
Recall that the range of the exponential function is (0, ∞).
Thus, 6x = 0 has no solution.
f) € Recall that the range of the exponential function is (0, ∞).
€
Thus, 5x = – 25 has no solution.
Now, suppose we want to solve 4x = 13. The problem with 13 is that it is not
a perfect power of four. We can make a conjecture that x should be
between 1 and 2 since 41 = 4 and 42 = 16 and 13 is between 4 and 16.
We could try some values for x
Try x = 1.5, we get 41.5 = 8
(too low)
1.75
Try x = 1.75, we get 4
≈ 11.3137... (too low)
1.875
Try x = 1.875, we get 4
≈ 13.4543... (too high)
1.8125
Try x = 1.8125, we get 4
≈ 12.3376... (too low)
Etc.
It will take quite a while to find x = 1.850219859... is the answer.
As we can see, this process is far too time consuming. We need to develop
a better method for solving this equation. Recall that if a function is one-toone, then it has an inverse function. Since an exponential function is oneto-one, then it has an inverse function. We call this inverse function a
logarithmic function.
167
Definition of a Logarithmic Function
For x > 0 and a > 0 and a ≠ 1, then
y = f(x) = loga (x) if and only if x = ay.
This is called the logarithmic function with base a and x is called the
argument. Note, the domain of loga (x) is (0, ∞) and the range is (– ∞, ∞).
Think of the logarithmic functions as asking the question:
"By what power f(x) do we have to raise the base a to get x?"
Argument = BasePower
Power = logBase(Argument)
Rewrite each exponential function as an logarithmic function:
Ex. 2a
3.53 = r
Ex. 2b
ew = 8
Ex. 2c
b6 = 56
Solution:
a)
Think "what power 3 do we have to raise 3.5 to get r?"
3 = log3.5(r)
b)
Think "what power w do we have to raise e to get 8?"
w = loge(8)
c)
Think "what power 6 do we have to raise b to get 56?"
6 = logb(56)
Rewrite each logarithmic function as an exponential function:
Ex. 3a
log12(144) = p
Ex. 3b
logc(1000) = 3
Ex. 3c
log4(x) = – 3
Solution:
a)
With log12(144) = p, 12 is the base, p is the power and 144 is
the “answer”. Thus, 12p = 144.
b)
With logc(1000) = 3, c is the base, 3 is the power and 1000
is the “answer”. Thus, c3 = 1000.
c)
With log4(x) = – 3, 4 is the base, – 3 is the power and
x is the “answer”. Thus, 4 – 3 = x.
Objective #2:
Evaluating Logarithmic Functions
Find the following:
Ex. 4a log10(100)
Ex. 4b log2( 5 2 )
Ex. 4c log3(
Ex. 4d log5 (5)
Ex. 4e log7 (1)
Ex. 4f log2 (0)
€
€
1)
81
168
Solution:
a)
For log10(100), ask the question: "By what power f(x) do we
have to raise the base 10 to get 100?" The answer is 2.
Thus, since 102 = 100, then log10(100) = 2.
b)
For log2( 5 2 ), ask the question: "By what power f(x) do we
have to raise the base 2 to get
c)
d)
5
2 = 21/5?" The answer is 1/5.
Thus, since 21/5 = 5 2 , then log2( 5 2 ) = 1/5.
€
For log3( 1 ), ask the question: "By what power f(x) do we have
81
€
to raise the base 3 to get 1 ?" Since 1 = 81 – 1 = (34) – 1 = 3 – 4,
€
€81
81
–4
the answer is – 4. Thus, since 3 = 1 , then log3( 1 ) = – 4.
81
81
€
For log5(5), ask the question: "By what power f(x) do we have
to raise the base€5 to get 5?"€The answer is 1.
Thus, since 51 = 5, then log5€
(5) = 1.
€
e)
For log7(1), ask the question: "By what power f(x) do we have
to raise the base 7 to get 1?" The answer is 0.
Thus, since 70 = 1, then log7(1) = 0.
f)
The domain of the logarithmic function is (0, ∞). Thus, log2 (0)
is undefined.
Objective #3:
Determining the Domain of a Logarithmic Function.
Since the logarithmic function is the inverse function of the exponential
function, the domain of a logarithmic function is equal to the range of the
exponential function, which is (0, ∞). In other words, for a logarithmic
function to be defined, the argument has to be greater than zero.
Find the domain of the following:
Ex. 5a
f(x) = log3(2 – 3x)
Solution:
a)
The argument has to
be greater than zero.
2 – 3x > 0
2
x<
3
The domain is (– ∞,
€
€
2
3
).
Ex. 5b g(x) = loge(11x + 30.25)
Solution:
b)
The argument has to
be greater than zero.
11x + 30.25 > 0
x > – 2.75
The domain is (– 2.75, ∞).
169
Ex. 5c
g(x) = log0.5
( x−3
x+2 )
Solution:
Solution:
x−3
x+2
c)
h(x) = log4(4 – x2)
Ex. 5d
>0
d)
4 – x2 > 0
3 is €
a zero.
(2 – x)(2 + x) > 0
x = – 2 is a V.A.
– 2 and 2 are zeros
We will need a sign chart:
We will need a sign chart:
€
Int: (– ∞, – 2)
(– 2, 3) (3, ∞) Int: (– ∞, – 2) (– 2, 2)
(2, ∞)
Val.
–3
0
4
Val.
–3
0
3
1
f(x) f(– 3) = 6 f(0) = – 1.5 f(4) =
f(x) f(– 3) = – 5 f(0) = 4 f(3) = – 5
Sign
+
–
6
+
Sign
–
+
Thus, the domain is
(– ∞, – 2) U (3,€∞)
Objective #4:
–
Thus, the domain is
(– 2, 2).
Graphing Logarithmic Functions.
Since logarithmic functions are inverse functions of exponential function,
then the graph of the logarithmic function and the graph of the exponential
function should be symmetrical to the line y = x.
a>1
0<a<1
4
4
ax
3
-4
-3
-2
-1
ax
3
2
2
1
1
0
0
-1
-2
0
1
2
loga(x)
3
4
-4
-3
-2
-1
-1
-2
-3
-3
-4
-4
0
1
2
3
loga(x)
Let’s verify these results by making a table of values and plotting some
points.
4
170
Make a table of values and graph the following:
Ex. 6a
f(x) = log2(x)
Solution:
a)
x
log2(x)
1
–3
8
1
4
1
2
€
€
b)
–2
–1
1
2
4
8
€
Ex. 6b
€
0
1
2
3
€
€
4
g(x) = log3(x)
x
1
27
1
9
1
3
log3(x)
–3
a)
1
3
9
27
f(x) = log2(x)
1
0
–1
€
0
1
2
3
g(x) = log3(x)
0
1
2
3
4
5
6
7
-1 0
-1
-2
-2
-3
-3
-4
-4
4
c)
h(x) = log4(x)
3
2
1
0
-4
–2
1
4
16
64
€
b) 3
1
-3
€
0
1
2
3
log4(x)
–3
1
64
1
16
1
4
–1
2
-2
x
–2
2
-1 0
-1
c)
4
3
-1 0
-1
Ex. 6c h(x) = log4(x)
1
2
3
4
5
6
7
1
2
3
4
5
Notice that these
functions have the
same basic shape
though as the base
gets bigger, the
graph gets "steeper."
In general, if a > 1,
loga(x) will have this
shape.
6
7
171
Make a table of values and graph the following:
Ex. 7a f(x) = log1/2(x) Ex. 7b g(x) = log1/3(x) Ex. 7c h(x) = log1/4(x)
Solution:
a)
x
8
4
2
1
1
2
1
4
1
8
€
€a)
€
log1/2(x)
–3
–2
–1
0
1
b)
x
27
9
3
1
1
3
1
9
1
27
2
3
€
4
f(x) = log1/2€
(x)
3
c)
2
3
4
5
6
7
-3
-4
-4
4
h(x) = log1/4(x)
3
2
1
0
-4
€
-1 0
-1
-3
-3
3
0
1
-2
-2
2
€ g(x) = log1/3(x)
3
-2
-1 0
-1
1
4
1
16
1
64
€
4
log1/4(x)
–3
–2
–1
0
1
1
0
-1 0
-1
x
64
16
4
1
3
2
1
c)
2
b)
€
2
log1/3(x)
–3
–2
–1
0
1
1
2
3
4
5
6
7
1
2
3
4
5
6
Notice that these
functions have the
same basic shape
though as the base
gets bigger, the
graph gets "steeper."
In general, if 0 < a < 1,
loga(x) will have this
shape.
7
172
Now, let's summarize what we know about logarithmic functions:
The graph of loga(x), a > 1
The graph of loga(x), 0 < a < 1
(1, 0)
(1, 0)
Domain: (0, ∞)
Domain: (0, ∞)
Range: (– ∞, ∞)
Range: (– ∞, ∞)
X-Intercept: (1, 0)
X-intercept: (1, 0)
No y-intercept
No y-intercept
Increasing
Decreasing
x = 0 is a Vertical Asymptote
x = 0 is a Vertical Asymptote
One-to-One
One-to-One
The Common Logarithmic Function
log10 (x) is called the common logarithmic function. It is usually denoted
as log (x). Thus, if there is no base written, we will assume that it is the
common log. The log key on a scientific calculator is the common
logarithmic function. Thus, y = log(x) if and only if 10y = x.
Evaluate the following using a calculator:
Ex. 8a log(376)
Ex. 8b log(1.6 × 107)
Ex. 8c log(0.0075)
Solution:
a)
log(376) = 2.5751878... ≈ 2.5752
b)
log(1.6 × 107) = log(16,000,000) = 7.20411998... ≈ 7.2041
c)
log(0.0075) = – 2.12493873... ≈ – 2.1249
173
The Natural Logarithmic Function
loge (x) is called the natural logarithmic function. It is usually denoted as
ln(x). Thus, if there it is written ln, we will assume that it is the natural log.
The ln key on a scientific calculator is the natural logarithmic function.
Thus, y = ln(x) if and only if ey = x.
Evaluate the following using a calculator:
Ex. 9a ln(95)
Ex. 9b ln(0.08)
Ex. 9c
Solution:
a)
ln(95) = 4.55387689... ≈ 4.5539
b)
ln(0.08) = – 2.525728644... ≈ – 2.5257
c)
ln(4.5) = 1.504077396... ≈ 1.5041
ln(4.5)
Since both e and 10 are greater than 1, their graphs will like the first type of
log graphs we examined:
f(x) = log(x)
-1
f(x) = ln(x)
4
4
3
3
2
2
1
1
0
0
-1
0
1
2
3
4
5
6
7
-1
-1
-2
-2
-3
-3
-4
-4
0
1
2
3
4
5
6
7
Graph the following Logarithmic Functions:
Ex. 10a g(x) = ln(x – 2) + 3
Ex. 10b h(x) = – 2•log(x + 3)
Solution:
a) This is the graph of
the natural log function
shifted up 3 and right 2.
Solution:
b) This is the graph of the common
log function stretched by 2, reflected
across the x-axis and shifted 3 left.
174
7
4
6
3
5
2
4
1
3
0
-4
2
-3
-2
1
-1
0
1
2
3
4
-2
0
-1
-1
-3
0
1
2
Objective #5:
3
4
5
6
7
8
-4
Solving Logarithmic Equations.
To solve an equation involving logarithms, we need to isolate the logarithm
on one side of the equation and then rewrite the equation in exponential
form.
Solve the following:
Ex. 11a
5•ln(4x – 1) = 10
Ex. 11c
log9(27) = r
Solution:
a)
5•ln(4x – 1) = 10
ln(4x – 1) = 2
4x – 1 = e2
x=
e2 +1
4
logb(625) = 4
7e3x + 2.5 = 20
(Solve for the natural log)
(Rewrite as an exponential equation)
(Solve for x)
≈ 2.0973
The solution is {
b)
Ex. 11b
Ex. 11d
e2 +1
}.
4
logb(625) = 4
(Rewrite as an exponential equation)
4
625 = b
(Solve for b)
2
b = ± 25
b2 = 25
or
b2 = – 25
b=±5
or
b = ± 5i
But, the base is always a positive real number, so reject – 5
and ± 5i. The solution is {5}.
175
c)
d)
log9(27) = r
27 = 9r
33 = (32)r
33 = 32r
2r = 3
r = 1.5
The solution is {1.5}.
(Rewrite as an exponential equation)
(27 = 33 and 9 = 32)
7e3x + 2.5 = 20
7e3x = 17.5
e3x = 2.5
3x = ln(2.5)
ln(2.5)
x=
≈ 0.3054
(Solve for e3x)
3
The solution is {
ln(2.5)
3
(3x is 1-1, so the exponents are equal)
(Rewrite as a logarithmic equation)
}.
Solve the following:
Ex. 12
The Richter scale defines the magnitude of an earthquake to be
M = log
( II ) where I is the intensity of the earthquake and I
o
o
is
the minimum intensity of an earthquake.
a)
If an earthquake has an intensity of 1,000,000Io, find the
Richter scale rating.
€
b)
If the Richter scale rating of an earthquake is 8, find how
many times stronger that earthquake is compared to the
earthquake in part a.
Solution:
a)
Replace I by 1,000,000Io and evaluate:
M = log
( 1000000I
)
I
o
o
= log 1000000 = 6
Thus, the earthquake would have a magnitude of 6 on the
Richter scale.
b)
€
Since M = 8, then log
( II ) = 8. Rewrite the equation as an
exponential equation:
I
Io
o
= 108 = 100,000,000.
So, I = 100,000,000Io which is 100 times stronger than the
earthquake is €
part a.
€
176
Find a function that might fit the graph below.
Ex. 13
4
Ex. 14
5
3
4
2
3
1
0
-4 -3 -2 -1
-1 0 1 2 3 4 5 6 7
-2
2
1
0
-3
-1 0 1 2 3 4 5 6 7 8 9 10 11 12
-4
-2
-5
-3
-6
-4
-7
-5
Solution:
The equation will be in the form:
f(x) = alogb(x – h). Since the vertical
asymptote is x = – 3, then h = – 3.
Since there is no reflection, a = 1.
Thus, f(x) = logb(x + 3). Plug in the
point (1, 2) to find b:
2 = logb(1 + 3)
2 = logb(4)
Write as an exponential equation:
b2 = 4
b=±2
Reject b = – 2. So, our function is:
f(x) = log2(x + 3)
You can verify that the point (5, 3)
satisfies the function.
Solution:
The equation will be in the form:
f(x) = alogb(x – h). Since the
vertical asymptote is x = 2, then
h = 2. Since there is no reflection,
a = 1. Thus, f(x) = logb(x – 2). Plug
in the point (5, – 1) to find b:
– 1 = logb(5 – 2)
– 1 = logb(3)
Write as an exponential equation:
b–1 = 3
b = 1/3
So, our function is
f(x) = log1/3(x – 2)
You can then verify that the point
(11, – 2) satisfies the function.